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# What is the 4-digit number in which the first digit is one-fifth of the last, and the second and third digits are the last digit multiplied by three?

We'll call this number $A B C D$ We have $A=\frac{1}{5}D$⇒$D=5A$ Since $D$ is divisible by 5, it is either 0 or 5 But it can't be 0, because then $A$ would be 0 So $D=5$ and $A=1$ Now I'm assuming that the second and third digits are adding to three times the last digit. So $B+C=3*5$⇒$B+C=15$ We have a number of options: $(B, C)=(6, 9)$ $(B, C)=(7, 8)$ $(B, C)=(8, 7)$ $(B, C)=(9, 6)$ So the possibilities are: 1695 1785 1875 1965 (If there were another detail we could narrow down the choices further, but otherwise we have all these options.)